1.1 Single mass systemConsider the

1.1 Single mass system
Consider the system of a single mass, supported by a spring and a dashpot, in which the damping is of a viscous character, see Figure 1.1. The
.........................................................................................................................................................................................
........................................................................................................................................................................................ ........ . ...
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
...........
....................................................................................................................................................... ................................................................
.....................................................................................................................................................................................
.........................................
.....................
.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
.
....................................................................................... ........................................................................................ ....................................................................................... ........................................................................................ ....................................................................................... .............
•..............................................................................
................ F
Figure 1.1: Mass supported by spring and damper.
spring and the damper form a connection between the mass and an immovable base
(for instance the earth).
According to Newton’s second law the equation of motion of the mass is
m
d2u
dt2 = P(t), (1.1)
where P(t) is the total force acting upon the mass m, and u is the displacement of
the mass.
It is now assumed that the total force P consists of an external force F(t), and
the reaction of a spring and a damper. In its simplest form a spring leads to a force
linearly proportional to the displacement u, and a damper leads to a response linearly
proportional to the velocity du/dt. If the spring constant is k and the viscosity of the
damper is c, the total force acting upon the mass is
P(t) = F(t) − ku − c
du
dt
. (1.2)
Thus the equation of motion for the system is
m
d2u
dt2 + c
du
dt
+ ku = F(t), (1.3)

結果 (繁體中文) 1:

1.1 單品質系統考慮系統的單的品質，支援由一個彈簧和阻尼器，在阻尼的粘度特性，參見圖 1.1。的................................................................................................................................................................................................................................................................................................................................................................................. ........ . ................................................................................................................................................................................................................................................................................................................................................................ ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ........................................................................................ ....................................................................................... ........................................................................................ ....................................................................................... .............•..............................................................................................F圖 1.1︰ 大眾支援的彈簧和阻尼器。彈簧和阻尼器之間形成連接的品質和不動產的基地（例如地球）。根據牛頓第二定律的質運動方程是md2udt2 = P(t)，(1.1)其中 P(t) 是總力在品質為 m，而你是的位移大眾。現在假定總力 P 由外力 f （t），和彈簧和阻尼器的反應。在其最簡單的形式春天導致一種力量位移 u 和阻尼器成正比線性線性導致回應速度 du/dt 成正比。如果彈簧常數 k 和粘度阻尼器是 c，被大眾的總作用力是P(t) = f （t) − ku − c杜dt.(1.2)因此該系統方程是運動的md2udt2 + c杜dt+ ku = f （t），(1.3)

結果 (繁體中文) 2:

1.1單質量系統
考慮一個單塊的系統中，由一個彈簧和一個緩衝器，在其中阻尼粘性性質的支持，參見圖1.1。該
................................................. .................................................. .................................................. ....................................
.............. .................................................. .................................................. .................................................. .................... ........。... ...
...........
...........
...........
...........
... ... ........
...........
...........
...........
......... ..
...........
...........
...........
...........
.... .......
...........
...........
...........
.......... 。
................................................. .................................................. .................................................. .. ................................................ ................
.................................. .................................................. .................................................. ..............................
... ......................................
............ .........
......................................... .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. ..............................
。
................... .................................................. .................. ................................ .................................................. ...... ........................................... ........................................... ....... .................................................. ............................... ................... .................................................. .................. .............
•.................. .................................................. ..........
................˚F
圖1.1：由彈簧和阻尼群眾支持
彈簧和阻尼器組成的質量和不動產之間的連接基
（例如地球）。
根據牛頓第二定律質量的運動方程是
米
D2U
DT2 = P（T），（1.1）
其中P（t）是作用在質量m的總力，和u是的位移
的質量。
現在假設，總力P由外力F（t）和
一個彈簧和阻尼器的反應。在其最簡單的形式的彈簧導致的力
線性比例的位移u和風門導致線性的響應
正比於速度的du / dt的。如果彈簧常數是k和粘度
阻尼為c，作用在質量的總力是
P（t）的= F（T） -ク- Visual C
杜
dt的
。（1.2）
因此，對系統的運動方程是
米
D2U
DT2 + C
杜
DT
+ク= F（T），（1.3）

結果 (繁體中文) 3:

1.1單質量系統考慮一個單質量系統，支持的一個彈簧和阻尼器的阻尼是粘性特徵，見圖1.1。這個…………………………………………………………………………………………………………………………………………………………………..………………………………………………………………………………………………………………………………………………………………….……..的…………..………..………..………..………..………..………..………..………..………..………..………..………..………..………..………..………..………..…………………………………………………………………………………………………………………………………….……………………………………………………….……………………………………………………………………………………………………………………………………………………………….…………………………………..………………….................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................。………………………………………………………………………………………………………………………………………………………….………………………………………………………………………………………………………………………………………………………….……………………………………………………………………………………….•………………………………………………………………………………….F圖1.1：彈簧和阻尼器的質量。彈簧和阻尼器之間形成一個質量和一個不可移動的基地之間的連接（例如地球）。根據牛頓的第二定律，質量的運動方程是MD2UDT2 = P（t），（1.1）其中P（t）是作用於質量m的總力，U是位移福斯。現在假定，總力P由一個外部力F（t），和一個彈簧和一個阻尼器的反應。在它最簡單的形式一個彈簧導致一個力線性成比例的位移U，和一個阻尼器的響應線性對速度的du / dt比例。如果彈簧常數k和粘度的阻尼器是C，作用於質量的總力是P（t）= F（t）−Ku−C杜DT。（1.2）囙此，該系統的運動方程MD2UDT2 + C杜DT+庫= F（t），（1.3）